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3 years ago

A 860 kΩ resistor has 34 μA of current. What is the supply voltage for this electric circuit?

Engineering

2 answers:

Juliette [100K]3 years ago

8 0

Explanation:

first changing kilo ohm to ohm

860000 = 860 kΩ

and change 34 micro ampare to ampare

34 μA=3.4×10^-5

recalling the equation V=I*R

V= 3.4×10^-5×860000

v=29.24

deff fn [24]3 years ago

6 0

Answer:

v=IR

V= 860 x 10^3 x 34 x 10^-6

V=29240 x 10^-3

V=29.24 V

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aniked [119]

The resources and instructions that should be used for the procedures of performing PMCS are:

Operator's manuals
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Fording kit
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PMCS is an acronym for preventive maintenance checks and services and it can be defined as the maintenance, checks, and services that are typically performed before, during, and after the use of any type of military equipment such as:

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Basically, the resources and instructions that should be used for the procedures of performing PMCS are:

Operator's manuals
Safety cautions and warnings.
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Read more on PMCS here: brainly.com/question/15720250

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1 year ago

An insulated piston-cylinder device contains 0.15 of saturated refrigerant-134a vapor at 0.8 MPa pressure. The refrigerant is no

Stells [14]

Answer:

Assumption:

1. The kinetic and potential energy changes are negligible

2. The cylinder is well insulated and thus heat transfer is negligible.

3. The thermal energy stored in the cylinder itself is negligible.

4. The process is stated to be reversible

Analysis:

a. This is reversible adiabatic(i.e isentropic) process and thus $s_{1} =s_{2}$

From the refrigerant table A11-A13

$P_{1} =0.8MPa$ $\left \{ {{ {{v_{1}=v_{g} @0.8MPa =0.025645 m^{3/}/kg } } \atop { {{u_{1}=u_{g} @0.8MPa =246.82 kJ/kg } - also {{s_{1}=s_{g} @0.8MPa =0.91853 kJ/kgK } } \right.$

sat vapor

m= $\frac{V}{v_{1} } =\frac{0.15}{0.025645} =5.8491 kg\\and \\\\P_{2} =0.2MPa \left \{ {{x_{2} =\frac{s_{2} -s_{f} }{s_{fg }}=\frac{0.91853-0.15449}{0.78339} = 0.9753 \atop {u_{2} =u_{f} +x_{2} }(u_{fg}) = 38.26+0.9753(186.25)= 38.26+181.65 =219.9kJ/kg \right. \\s_{1} = s_{2}$

$T_{2} =T_{sat @ 0.2MPa} = -10.09^{o} C$

b.) We take the content of the cylinder as the sysytem.

This is a closed system since no mass leaves or enters.

Hence, the energy balance for adiabatic closed system can be expressed as:

$E_{in} - E_{out} =$ ΔE

$w_{b, out} =$ ΔU

$w_{b, out} =m([tex]u_{1} -u_{2$ )

$w_{b, out} =$ workdone during the isentropic process

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3 years ago

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Arisa [49]

Answer:

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Subsidies for farmers

Insurance for farmers

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Food assistance for economically challenged Americans (the largest portion of the Farm Bill)

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3 years ago

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Anna35 [415]

Answer:

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